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Let $k$ be a finite field of characteristic $p>0$ with $q$ elements. Denote $K=W(k)[1/p]$. Let $E$ be an elliptic curve over $W(k)$ with good reduction. Choose a lifting $\mathrm{Frob} \in \mathrm{Gal}(\bar K/K)$ of the $q$-th power map on $\overline{k}$ such that $\chi(\mathrm{Frob})=1$, here $\chi$ is the $p$-adic cyclotomic character. Recall that the trace of $\mathrm{Frob}$ on $\ell$-adic Tate module does not depend on the choice of the prime number $\ell(\neq p)$ and is equal to $ a_q := 1 + \# k - \# E(k) $.

Question: Is the trace of $\mathrm{Frob}$ on the $p$-adic Tate module still equal to $a_q$?

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  • $\begingroup$ Yes; as the corresponding representation will be crystalline. $\endgroup$
    – Stanley Yao Xiao
    Commented Jul 16, 2019 at 11:58
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    $\begingroup$ The $\bmod $ reduction is an isomorphism $E[\ell^\infty] \to \widetilde{E}[\ell^\infty]$ so the matrices in $GL_2(\Bbb{Z}_\ell)$ of $Frob|_{E[\ell^\infty]}$ and its reduction are the same, the OP is asking what does the matrix of $Frob |_{E[\ell^p]}\in Aut(E[p^\infty])\cong GL_2(\Bbb{Z}_p)$ look like and if its trace is related to $tr(Frob|_{E[\ell^\infty]})$ $\endgroup$
    – reuns
    Commented Jul 16, 2019 at 12:07
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    $\begingroup$ Why do you even expect the trace of $\mathrm{Frob}$ to be independent of the choice of lift? $\endgroup$
    – naf
    Commented Jul 16, 2019 at 12:14
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    $\begingroup$ If E is ordinary, then the action is upper triangularizable with a quotient determined by the reduction, and a sub determined by that quotient and the cyclotomic character, so the trace is determined by how the galois element acts on p^n ropts of unity, which presumably depends on the lift of element in galois but not on the choice of lift of elliptic curve $\endgroup$
    – Atticus Christensen
    Commented Jul 17, 2019 at 0:00
  • $\begingroup$ Thanks. You are right, the trace does depend on the choice of the lifting of the Frobenius map. Maybe, I need to consider those lifting $\mathrm{Frob}$ such that $\chi(Frob)=1$, here $\chi$ is the cyclotomic character. $\endgroup$
    – Yang
    Commented Jul 17, 2019 at 9:15

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